Consider the following compounds: $\mathrm{KO}{}_2$, $\mathrm{H}{}_2\mathrm{O}{}_2$ and $\mathrm{H}{}_2\mathrm{SO}{}_4$. The oxidation states of the underlined elements ($\mathrm{K}$ in $\mathrm{KO}{}_2$, $\mathrm{O}$ in $\mathrm{H}{}_2\mathrm{O}{}_2$, $\mathrm{S}$ in $\mathrm{H}{}_2\mathrm{SO}{}_4$) in them are, respectively,

Solution & Explanation

\textbf{Answer:} (c) $+1$, $-1$ and $+6$. \textbf{Solution:} $\mathrm{KO}{}_2$ is potassium superoxide, in which $\mathrm{K}$ is $+1$ (each $\mathrm{O}$ is $-1/2$). $\mathrm{H}{}_2\mathrm{O}{}_2$ is a peroxide, so $\mathrm{O}$ is $-1$. In $\mathrm{H}{}_2\mathrm{SO}{}_4$: $2(+1)+S+4(-2)=0$, giving $\mathrm{S}$ $=+6$. Hence the underlined elements are $+1$ ($\mathrm{K}$), $-1$ ($\mathrm{O}$ in peroxide) and $+6$ ($\mathrm{S}$), i.e. option (c).